8.1.1 Confidence Intervals

Suppose heights, in inches, of orangutans are normally distributed and have a known population standard deviation of 4 inches and an unknown population mean. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval for the population mean with a 95% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Round the final answer to two decimal places.

(54.04 , 57.96) use the Confidence interval equation. The point estimate is the sample mean, x¯, and the margin of error is margin of error = (z α/2)(σ/√n) Substituting the given values σ=4, n=16, and z α/2=1.96 for a confidence level of 95%, we have margin of error=(1.96)(4/√16) ≈(1.96)(1.0) ≈1.96 With x=56 and a margin of error of 1.96, the confidence interval is (56 − 1.96 , 56 + 1.96) (54.04,57.96). So we estimate with 95% confidence that the true population mean is between 54.04 and 57.96 inches.

Adult entrance fees to amusement parks in the United States are normally distributed with a population standard deviation of 2.5 dollars and an unknown population mean. A random sample of 22 entrance fees at different amusement parks is taken and results in a sample mean of 61 dollars. Find the margin of error for a 99% confidence interval for the population mean. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

1.37 The margin of error is given by margin of error = (z α/2)(σ/√n) Substituting the given values σ=2.5, n=22, and z α/2 =2.576 for a confidence level of 99%, we have margin of error =(2.576)(2.5/√22) ≈(2.576)(0.53) ≈1.37

A bank offers auto loans to qualified customers. The amount of the loans are normally distributed and have a known population standard deviation of 4 thousand dollars and an unknown population mean. A random sample of 22 loans is taken and gives a sample mean of 42 thousand dollars. Find the margin of error for the confidence interval for the population mean with a 90% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

1.40 σ=4, n=22, and z α/2 = 1.645 for a confidence level of 90%, we have margin of error: =(1.645)(4/√22) ≈(1.645)(0.852) ≈1.40

The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 6 inches and an unknown population mean. If a random sample of 18 snakes is taken and results in a sample mean of 61 inches, find the margin of error (ME) of the confidence interval with a 90% confidence level. Round your answer to three decimal places. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

2.326. We can use the formula to find the margin of error: ME=(z α/2)(σ/√n) We know that σ=6 and n=18. We are also given that the confidence level (CL) is 90%, or 0.9. So, we can calculate alpha (α). α=1 =CL=1 =0.9 =0.1 Since α=0.1, we know that α2 =0.12 =0.05 The value of z0.05 is 1.645. Now we can substitute the values into the formula to find the margin of error. ME =(z α/2)(σ/√n) =(1.645)(6/√18) ≈(1.645)(1.414) ≈2.326 So, the margin of error (ME) is 2.326.

The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation of 425 dollars and an unknown population mean. A random sample of 22 sociologists is taken and gives a sample mean of 1520 dollars. Find the margin of error for the confidence interval for the population mean with a 98% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

210.76 values σ=425, n=22, and zα2=2.326 for a confidence level of 98%, we have margin of error =(2.326)(425/√22) ≈(2.326)(90.61) ≈210.76

In a recent survey, a random sample of 130 families were asked about whether they have a pet, and 67 reported that they have a pet. What value of z should be used to calculate a confidence interval with a 90% confidence level? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

25 In this formula, z = z α/2 = z0.025 =1.96 because the confidence level is 95%. σ=5 and ME =2. Therefore, n= z2σ2/ME2 =(1.96)^2(5)^2/2^2 ≈24.01 Use n=25 to ensure that the sample size is large enough.

The population standard deviation for the body weights for employees of a company is 10 pounds. If we want to be 95% confident that the sample mean is within 3 pounds of the true population mean, what is the minimum sample size that should be taken.

43 employees Confidence Level =0.950 StDev =10 Error =3 z-Value =1.960 Minimum Sample Size =4315

he population standard deviation for the total snowfalls per year in a city is 13 inches. If we want to be 95%confident that the sample mean is within 3 inches of the true population mean, what is the minimum sample size that should be taken?

73 snowfalls Confidence Level 0.950 StDev 13 Error 3 z-Value 1.960 Minimum Sample Size 7314.

The weights, in pounds, of dogs in a city are normally distributed with a population standard deviation of 2 pounds and an unknown population mean. A random sample of 16 dogs is taken and results in a sample mean of 28 pounds. Identify the parameters needed to calculate a confidence interval at the 90% confidence level. Then find the confidence interval. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Round the final answer to two decimal places.

Confidence intervals are written as (point estimate−margin of error, point estimate + margin of error) The point estimate is the sample mean, x¯, and the margin of error=(z α/2)(σ/√n) Substituting the given values σ=2, n=16, and z α/2=1.645 for a confidence level of 90%, margin of error =(1.645)(2/√16) ≈ (1.645)(0.5) ≈0.82 With x¯=28 and a margin of error of 0.82, the confidence interval is (28−0.82,28+0.82) =(27.18,28.82). We can estimate with 90% confidence that the true population mean weight of dogs in the city is between 27.18 and 28.82 pounds.

A researcher is trying to estimate the population mean for a certain set of data. The sample mean is x¯=45, and the margin of error for the mean is 9. Find the corresponding confidence interval for the mean.

We are given the sample mean x¯=45, and a margin of error of 9. So the confidence interval estimate, at the 99.7% confidence level, is (45−9,45+9), which simplifies to the confidence interval (36,54).

Suppose the finishing times for cyclists in a race are normally distributed. If the population standard deviation is 16 minutes, what minimum sample size is needed to be 95% confident that the sample mean is within 5 minutes of the true population mean? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

formula used z =z α/2 = z0.025 =1.96 because the confidence level is95%. σ= 16 and ME = 5. Therefore, n = z^2σ^2/ME^2 =(1.96)^2(16)^2/5^2 ≈39.34 Use n= 40 to ensure that the sample size is large enough.

Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

formula used z =z α/2 = z0.05 =1.645 because the confidence level is90%. σ= 3 and ME = 1. Therefore, n = z^2σ^2/ME^2 =(1.645)^2(3)^2/1^2 ≈ Use n= 40 to ensure that the sample size is large enough.

The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches. Identify the parameters needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

x = ​ 58 σ = ​ 8 n = ​ 25 z α/2​​ = ​ 2.576 (53.88, 62.12) the given values σ=8, n=25, and z α/2=2.576 for a confidence level of 99%, we have margin of error=(2.576)(8/√25) ≈ (2.576)(1.6) ≈4.12 With x¯=58 and a margin of error of 4.12, the confidence interval is (58−4.12,58+4.12) = (53.88 , 62.12). We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches.

Suppose heights of seasonal pine saplings are normally distributed and have a known population standard deviation of 17 millimeters and an unknown population mean. A random sample of 15 saplings is taken and gives a sample mean of 308 millimeters. Find the confidence interval for the population mean with a 90% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

σ=17 n=15 z α/2 =1.645 confidence level of 90%, we find the margin of error = (1.645)(17/√15) ≈ (1.645)(4.389) ≈7.22 x¯=308 subtract M.O.E = 7.22, the confidence interval: (308 − 7.22 , 308 + 7.22) = (300.78,315.22). So we estimate with 90% confidence that the true population mean is between 300.78 and 315.22 millimeters.

Find the confidence interval: (a) for the true mean height of American women, given a sample mean of x¯=64 inches, and margin of error 2.4 inches. (b) for the true proportion of voters who support an issue, given a sample proportion of p̂ =0.45, and margin of error 0.04.

(a) The point estimate is the sample mean x¯=64 inches, and the margin of error is 2.4 inches. The confidence interval is: (64−2.4,64+2.4)(61.6,66.4) (b) The point estimate is the sample proportion p̂ =0.45, and the margin of error is 0.04. The confidence interval is: (0.45−0.04,0.45+0.04)(0.41,0.49)

The population standard deviation for the typing speeds for secretaries is 4 words per minute. If we want to be 90% confident that the sample mean is within 1 word per minute of the true population mean, what is the minimum sample size that should be taken?

≈43.30